Question

In the circuit shown in the figure, each battery is of $5V$ and has an internal resistance of $0.2\Omega$. What is the reading of the ideal voltmeter $V$?

Answer 1

As the internal resistance of ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as
$\frac{1}{r_1}=\frac{1}{r}+\frac{1}{\infty }\Rightarrow r_1=r$
Now, as the given eight batteries are discharging in seris, i.e.,
$E_{eq}=8\times 5=40V;r_{eq}=8\times 0.2=1.6\Omega$
So, current in the circuit, $I=\frac{E_{eq}}{r_{eq}}=\frac{4/0}{1.6}=25A$
hence, potential difference across the required battery
$V=E-Ir=5-25\times 0.2V$